Question: Let $f(x)=\log_3(x-4x^2)$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1-8x}{(x-4x^2)\log_3(x)}$ (Choice B) B $\dfrac{1}{(x-4x^2)\ln(3)}$ (Choice C) C $\dfrac{1-8x}{(x-4x^2)\ln(3)}$ (Choice D) D $\dfrac{1-8x}{x-4x^2}$
Explanation: $f$ is a logarithmic function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=x-4x^2$, then $f(x)=\log_3\Bigl(u(x)\Bigr)$. $f'(x)$ can be found using the following identity: $\dfrac{d}{dx}\left[\log_3\Bigl(u(x)\Bigr)\right]=\dfrac{u'(x)}{\ln(3)u(x)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\log_3(x-4x^2) \\\\ &=\dfrac{d}{dx}\log_3\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=x-4x^2} \\\\ &=\dfrac{u'(x)}{\ln(3)u(x)} \\\\ &=\dfrac{1-8x}{\ln(3)(x-4x^2)}&&\gray{\text{Substitute }u(x)\text{ back}} \\\\ &=\dfrac{1-8x}{(x-4x^2)\ln(3)} \end{aligned}$ In conclusion, $f'(x)=\dfrac{1-8x}{(x-4x^2)\ln(3)}$.